Problem: Multiply the following complex numbers: $({-2+2i}) \cdot ({3+2i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-2+2i}) \cdot ({3+2i}) = $ $ ({-2} \cdot {3}) + ({-2} \cdot {2}i) + ({2}i \cdot {3}) + ({2}i \cdot {2}i) $ Then simplify the terms: $ (-6) + (-4i) + (6i) + (4 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -6 + (-4 + 6)i + 4i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -6 + (-4 + 6)i - 4 $ The result is simplified: $ (-6 - 4) + (2i) = -10+2i $